Q. 3 - Calculus

Question

Each of the integral expressions that follow represents the area of a region in the plane bounded by a function expressed in polar coordinates. Use the ideas from this section and from Chapter 9 to sketch the regions, and then evaluate each integral   

${\int }_0^{\frac{2\mathrm{\pi }}{3}}{\left(\frac{1}{2}+\cos \theta \right)}^{2}d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}{\left(\frac{1}{2}+\cos \theta \right)}^{2}d\theta$

Step-by-Step Solution

Answer

The value of integral is $\frac{\mathrm{\pi }}{4}$

Step 1 .Given information

Integral;

${\int }_0^{\frac{2\mathrm{\pi }}{3}}{\left(\frac{1}{2}+\cos \theta \right)}^{2}d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}{\left(\frac{1}{2}+\cos \theta \right)}^{2}d\theta$

Step 2. Plot the region:

In the given integral we can see that there is a subtraction occurs between integral.

By comparing the given integral with the formula of area of region of the polar curve $r=f\left(\theta \right)$:

$\frac{1}{2}{\int }_a^b{r}^{2}d\theta$ 

We get $r=\sqrt{2}\left(\frac{1}{2}+\cos \theta \right)$

Now plot this curve in both given interval of $\theta$

Step 3. Simplify integral

$$\begin{gathered} {\int }_0^{\frac{2\mathrm{\pi }}{3}}{\left(\frac{1}{2}+\cos \theta \right)}^{2}d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}{\left(\frac{1}{2}+\cos \theta \right)}^{2}d\theta \\ ={\int }_0^{\frac{2\mathrm{\pi }}{3}}\left(\frac{1}{4}+{\cos }^2\theta +\cos \theta \right)d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}\left(\frac{1}{4}+{\cos }^2\theta +\cos \theta \right)d\theta \\ ={\int }_0^{\frac{2\mathrm{\pi }}{3}}\left(\frac{1}{4}+\frac{1+\cos 2\theta }{2}+\cos \theta \right)d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}\left(\frac{1}{4}+\frac{1+\cos 2\theta }{2}+\cos \theta \right)d\theta \\ ={\int }_0^{\frac{2\mathrm{\pi }}{3}}\left(\frac{1}{4}+\frac{1}{2}+\frac{\cos 2\theta }{2}+\cos \theta \right)d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}\left(\frac{1}{4}+\frac{1}{2}+\frac{\cos 2\theta }{2}+\cos \theta \right)d\theta \\ ={\int }_0^{\frac{2\mathrm{\pi }}{3}}\left(\frac{3}{4}+\frac{\cos 2\theta }{2}+\cos \theta \right)d\theta -{\int }_{\pi }^{\frac{4\mathrm{\pi }}{3}}\left(\frac{3}{4}+\frac{\cos 2\theta }{2}+\cos \theta \right)d\theta \end{gathered}$$

Step 4. Solve integral

Now integrall can be solved as

$$\begin{gathered} {\left[\frac{3}{4}\theta +\frac{1}{2}\frac{sin2\theta }{2}+\sin \theta \right]}_0^{\frac{2\mathrm{\pi }}{3}}-{\left[\frac{3}{4}\theta +\frac{1}{2}\frac{sin2\theta }{2}+\sin \theta \right]}_{\mathrm{\pi }}^{\frac{4\mathrm{\pi }}{3}} \\ =\left[\frac{3\times {\displaystyle \frac{2\mathrm{\pi }}{3}}}{4}-\frac{1}{4}\sin \left(\frac{4\mathrm{\pi }}{3}\right)+\sin \left(\frac{2\mathrm{\pi }}{3}\right)-0\right]-\left[\frac{3\times {\displaystyle \frac{4\mathrm{\pi }}{3}}}{4}-\frac{1}{4}\sin \left(\frac{8\mathrm{\pi }}{3}\right)+\sin \left(\frac{4\mathrm{\pi }}{3}\right)-\left(\frac{3}{4}\mathrm{\pi }+\frac{1}{4}\sin \left(2\mathrm{\pi }\right)+\mathrm{sin\pi }\right)\right] \\ =\left(\frac{\mathrm{\pi }}{2}-\frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{2}\right)-\left(\mathrm{\pi }-\frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{2}-\frac{3}{4}\mathrm{\pi }-0-0\right) \\ =\frac{\mathrm{\pi }}{2}-\frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{2}-\mathrm{\pi }+\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{2}+\frac{3}{4}\mathrm{\pi } \\ =\frac{3\mathrm{\pi }-2\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$